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RE: st: RE: How to define shortest possible period with 95% of observations


From   "Nick Cox" <n.j.cox@durham.ac.uk>
To   <statalist@hsphsun2.harvard.edu>
Subject   RE: st: RE: How to define shortest possible period with 95% of observations
Date   Wed, 12 May 2010 20:13:38 +0100

What you outline is not what I was recommending. It's an awkward
half-way house. 

But in terms of your new question, see 

SJ-6-4  dm0025  . . . . . . . . . .  Stata tip 36: Which observations?
Erratum
        . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  N.
J. Cox
        Q4/06   SJ 6(4):596                              (no commands)
        correction of example code for Stata tip 36

SJ-6-3  dm0025  . . . . . . . . . . . . . .  Stata tip 36: Which
observations?
        . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  N.
J. Cox
        Q3/06   SJ 6(3):430--432                                 (no
commands)
        tip for identifying which observations satisfy some
        specified condition

Nick 
n.j.cox@durham.ac.uk 

Daniel Mueller

Thanks.

In the code below I did my best to think in fire years by defining the 
peak fire day (these are always unique) as the middle of the year. Then 
I'd like to place the maximum into a local macro (where I fail miserably

in line 2 as Steve rightly pointed out):

qui su no_fire_day
loc yearstart = Day[r(max)] - 183
loc yearend = `yearstart' + 365

My simple question is how I can place the Day where no_fires_day = 
r(max) into a local macro?

(I ignore leap years for the sake of simplicity..)


Nick Cox wrote on 5/12/2010 10:44 PM:
> Without looking at this in detail, it seems to me that you might
benefit
> from thinking in terms of fire years, rather than calendar years,
> starting on some day other than January 1.  After all, all sorts of
> different sciences, not to mention religions, have years that don't
> coincide with conventional Western calendar years: fiscal years, water
> years, academic years, etc., etc.
>
> Several pertinent -egen- functions are included in -egenmore- on SSC.
>
> In other words, define the time scale in terms of those fire years;
then
> Robert's code will probably not need any complicated adjustments.
>
> Nick
> n.j.cox@durham.ac.uk
>
> Daniel Mueller
>
> Robert, this works like charm!!! Thanks a bunch for this neat code.
Also
>
> thanks to Nick for pointing me to -shorth- which I will certainly
> explore in more detail after having sipped through the extensive
> reference list.
>
> Using Roberts code I can seamlessly loop over the nine years of data
and
>
> generate the shortest fire season per year with 95% of obs. The
results
> suggested an additional complication.. For some subsets the shortest
> possible period likely starts a couple of days before Jan 1st, at the
> end of the preceding year.
>
> I tweaked Roberts code a little to loop over years and defined the
> middle of a year as the peak fire day. The code runs through, yet sets
> the start of the fire season for some subsets to Jan 1st, while my
> educated guess is that it should be somewhere around mid to end of
> December. Something went wrong, but I can't spot the glitch in the
code
> below. Can someone please help?
>
> Thanks a lot in advance and best regards,
> Daniel
>
>
> *** start
> forv y = `yearfirst'/`yearlast' {
>
> * keep previous year
>    if `y' != `yearfirst' {
>     keep if Year == `y' | Year == (`y'-1)
>    }
>    bys Day: g no_fire_day = _N
>    qui su no_fire_day
>
> * define year to start 183 days before peak fire day
>    loc yearstart = Day[r(max)] - 183
>    loc yearend = `yearstart' + 365
>    keep if Day>  `yearstart'&  Day<  `yearend' // or with
egen->rotate?
>    bys Day: keep if _n == _N
>    g nobs = _n
>
> * the target is a continuous run that includes 95% of all fires
>    sum no_fire_day, meanonly
>    scalar target = .95 * r(sum)
>
>    scalar shortlen = .
>    gen arun = .
>    gen bestrun = .
>
>    * at each pass, create a run that starts at nobs == `i'
>    * and identify the nobs where the number of fires>= 95%
>    local more 1
>    local i = 0
>    while `more' {
>     local i = `i' + 1
>     qui replace arun = sum(no_fire_day * (nobs>= `i'))
>     sum nobs if arun>= target, meanonly
>     if r(N) == 0 local more 0
>     else if (Day[r(min)] - Day[`i'])<  shortlen {
>      scalar shortlen = Day[r(min)] - Day[`i']
>      qui replace bestrun = arun
>      qui replace bestrun = . if nobs>  r(min) | nobs<  `i'
>     }
>    }
>    qui drop if bestrun == .
>    drop bestrun arun
>    save fires_`y', replace
> }
> *** end
>
>
>
>
>
> Robert Picard wrote on 5/11/2010 3:28 AM:
>> Here is how I would approach this problem. I would do each year
>> separately; it could be done all at once but it would complicate the
>> code unnecessarily. If the fire data is one observation per fire, I
>> would -collapse- it to one observation per day. Each observation
would
>> contain the number of fires that day. The following code will
identify
>> the first instance of the shortest run of days that includes 95% of
>> fires for the year.
>>
>> Note that the following code will work, even if there are days
without
>> fires (and thus no observation for that day).
>>
>> *--------------------------- begin example -----------------------
>> version 11
>>
>> * daily fire counts; with some days without fires
>> clear all
>> set seed 123
>> set obs 365
>> gen day = _n
>> drop if uniform()<   .1
>> gen nobs = _n
>> gen nfires = round(uniform() * 10)
>>
>> * the target is a continuous run that includes 95% of all fires
>> sum nfires, meanonly
>> scalar target = .95 * r(sum)
>> dis target
>>
>> scalar shortlen = .
>> gen arun = .
>> gen bestrun = .
>>
>> * at each pass, create a run that starts at nobs == `i'
>> * and identify the nobs where the number of fires>= 95%
>> local more 1
>> local i 0
>> while `more' {
>> 	local i = `i' + 1
>> 	qui replace arun = sum(nfires * (nobs>=`i'))
>> 	sum nobs if arun>= target, meanonly
>> 	if r(N) == 0 local more 0
>> 	else if (day[r(min)] - day[`i'])<   shortlen {
>> 		scalar shortlen = day[r(min)] - day[`i']
>> 		qui replace bestrun = arun
>> 		qui replace bestrun = . if nobs>   r(min) | nobs<   `i'
>> 	}
>> }
>>
>> *--------------------- end example --------------------------
>>
>>
>> Hope this help,
>>
>> Robert
>>
>> On Mon, May 10, 2010 at 6:19 AM, Nick Cox<n.j.cox@durham.ac.uk>
> wrote:
>>> I don't think any trick is possible unless you know in advance the
>>> precise distribution, e.g. that it is Gaussian, or exponential, or
>>> whatever, which here is not the case.
>>>
>>> So, you need to look at all the possibilities from the interval
> starting
>>> at the minimum to the interval starting at the 5% point of the fire
>>> number distribution in each year.
>>>
>>> However, this may all be achievable using -shorth- (SSC). Look at
the
>>> -proportion()- option, but you would need to -expand- first to get a
>>> separate observation for each fire. If that's not practicable, look
>>> inside the code of -shorth- to get ideas on how to proceed. Note
that
> no
>>> looping is necessary: the whole problem will reduce to use of -by:-
> and
>>> subscripts.
>>>
>>> Nick
>>> n.j.cox@durham.ac.uk
>>>
>>> Daniel Mueller
>>>
>>> I have a strongly unbalanced panel with 100,000 observations (=fire
>>> occurrences per day) that contain between none (no fire) and 3,000
> fires
>>>
>>> per day for 8 years. The fire events peak in March and April with
> about
>>> 85-90% of the yearly total.
>>>
>>> My question is how I can define the shortest possible continuous
> period
>>> of days for each year that contains 95% of all yearly fires. The
> length
>>> and width of the periods may slightly differ across the years due to
>>> climate and other parameters.
>>>
>>> I am sure there is a neat trick in Stata for this, yet I have not
>>> spotted it. Any suggestions would be appreciated.
>
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>


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